Solve {left(-1/7right)}^2-4*2/5*(-5/12)=

$${ \left(- \frac{ 1 }{ 7 } \right) }^{ 2 } -4 \times \frac{ 2 }{ 5 } \times (- \frac{ 5 }{ 12 } )=$$

Evaluate

$$\frac{101}{147}\approx 0.68707483$$

Solution Steps

Calculate -\frac{1}{7} to the power of 2 and get \frac{1}{49}.

$$\frac{1}{49}-4\times \frac{2}{5}\left(-\frac{5}{12}\right)$$

Express 4\times \frac{2}{5} as a single fraction.

$$\frac{1}{49}-\frac{4\times 2}{5}\left(-\frac{5}{12}\right)$$

Multiply 4 and 2 to get 8.

$$\frac{1}{49}-\frac{8}{5}\left(-\frac{5}{12}\right)$$

Multiply \frac{8}{5} times -\frac{5}{12} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{49}-\frac{8\left(-5\right)}{5\times 12}$$

Do the multiplications in the fraction \frac{8\left(-5\right)}{5\times 12}.

$$\frac{1}{49}-\frac{-40}{60}$$

Reduce the fraction \frac{-40}{60} to lowest terms by extracting and canceling out 20.

$$\frac{1}{49}-\left(-\frac{2}{3}\right)$$

The opposite of -\frac{2}{3} is \frac{2}{3}.

$$\frac{1}{49}+\frac{2}{3}$$

Least common multiple of 49 and 3 is 147. Convert \frac{1}{49} and \frac{2}{3} to fractions with denominator 147.

$$\frac{3}{147}+\frac{98}{147}$$

Since \frac{3}{147} and \frac{98}{147} have the same denominator, add them by adding their numerators.

$$\frac{3+98}{147}$$

Add 3 and 98 to get 101.

$$\frac{101}{147}$$

Factor

$$\frac{101}{3 \cdot 7 ^ {2}} = 0.6870748299319728$$